Objective: Find the area of any triangle given at least ...€¦ · Objective: Find the area of any...

Objective: Find the area of any triangle given at

least three pieces of information.

Process: Derive several formulas to allow use

of given information (to avoid rounding errors).

By: Anthony E. Davis

Summer 2003

Where to start?

c b

a

B Ca

b

Ca

Not Quite Sure? - Try Some Practice Problems.

If you have two sides

and the included angle

then click here.

If you have two angles

and the included side

then click here.

If you have all three

sides then click here.

b

Ca

Two sides and the Included angle

See derivation.

Look at formula.

Try an example.

Return to choices.

B Ca

Two angles and the included side

See derivation.

Look at formula.

Try an example.

Return to choices.

c b

a

All three sides

Look at formula.

Try an example.

Return to choices.

Not Quite Sure?

•Draw a triangle.

•Label the information given.

•Match this triangle with one of the

three shown. Remember all triangles

and all variables are arbitrarily drawn

so rotation may be necessary.

•Return to choices

Derivation

•We know Area=1/2(base)(height).

•Let a represent the base.

•Using right triangle trigonometry, sin C = h/b

•Solve for h, h = b sin C.

•Replace values in area formula: Area = 1/2 a (bsin C)

•Hence the Area given two sides and the included angle

is any of the following:Area = 1/2 ab sin C

Area = 1/2 bc sin A

Area = 1/2 ac sin B

A

c b

B C

a

h

Formula

for

Two Sides and the Included

Angle

Area 1

2absinC

Example

Find the area of ∆DEF, if d = 3 cm, e = 8 cm

and F = 35°. Round to the nearest hundredth.

Area 1

2desinF

Area 1

2(3)(8)sin35

Area 6.88cm2

Derivation

•We know Area = 1/2 (base)(height)

•Let a represent the base

•Find the third angle by subtracting the two known

from 180.

•From right triangle trigonometry, sin C = h/b

•Solve for h, h = b sin C

•Replacing values, Area = 1/2 a (bsin C).

A

c b

B C

a

h

Derivation (cont.)

•However we are only given one side, so we need to

substitute the ‘a’ or ‘b’ out. (Let say the ‘b’).

•From the Law of Sines, sin A/a = sin B/b. We know

A, B and ‘a’ so we will solve for ‘b’.

•Solve for ‘b’, b = (a sin B)/(sin A)

•Replace, Area = 1/2 a ((a sin B)/(sin A)) sin C

•Thus we have the following

A

c b

B C

a

h

Area 1

2

a2 sinBsinC

sinA

Formula

for

Two Angles and the Included

Side

Area 1

2

a2 sinBsinC

sinA

Example

Find the area of ∆CAB if b = 7 ft., C = 42º, and B = 28º. Round your answer to the nearest tenth.

Angle A = 180- B - C

A 180 28

42

A 110 Area

1

2

b2sinAsinC

sinB

Area 1

2

(7)2 sin110sin 42

sin 28

Area 32.8 ft2

Formula

for

All Three Sides

(Heron’s Formula)

( )( )( )

a+b+c where S (semiperimeter) =

2

Area S S a S b S c

Example

Find the area of an equilateral triangle having legs of

length 3.2 mm. Round your answer to two decimal

places.

S a b c

2, in this case a = b = c = 3.2

S =3.2 + 3.2 + 3.2

2

S 4.8mm Area (4.8)(4.8 3.2)(4.8 3.2)(4.8 3.2)

Area 19.6608

Area 4.43mm2

Practice Problems

Directions: Find the area of each triangle using the given

information. Round only your final answer to the nearest

tenth. You may click on the question to see the solution.

1. ABC, a = 3, b = 2, C = 24

2. CBH, h = 3, C = 49, H = 24

3. DKP, d = 6, k =14, p = 24

4. HYM , h = 4, M = 18, H = 61

5. DGH, d = 7, g = 9, h = 4

6. CFV, c = 31, F = 27, v = 26

Answer

Answer

Answer

Answer

Answer

Answer

Answer #1:

ABC, a = 3, b = 2, C = 24

Area 1

2absinC

Area 1

2(3)(2)sin 24

Area 1.2units 2

Return to problems.

Answer #2

CBH, h = 3, C= 49, H = 24

B 180 49 24

B 107Area

1

2

h2sinCsinB

sinH

Area 1

2

(3)2 sin 49sin107

sin 24

Area 8.0units 2

Return to problems.

Answer #3

DKP, d = 6, k = 14, p = 24

S d k p

2

S 6 14 24

2

S 22units

Area S(S d)(S k)(S p)

Area 22(22 6)(22 14)(22 24)

Area 5632

Cannot take square root of negative number.

These three sides (6, 14, 24) do not form a triangle.

Remember the two smaller sides must add to more

than the third side.

Return to problems.

Answer #4

HYM, h = 4, M =18, H = 61

Y 180 18 61

Y 101

Area 1

2

h2sinMsinY

sinH

Area 1

2

(4)2 sin18sin101

sin61

Area 2.8units 2

Return to problems.

Answer #5

DGH, d = 7, g = 9, h = 4

S d g h

2

S 7 9 4

2

S 10units Area S(S d)(S g)(S h)

Area 10(10 7)(10 9)(10 4)

Area 180

Area 13.4units 2

Return to problems.

Answer #6

CFV, c = 31, F = 27, v= 26

Area 1

2cv sinF

Area 1

2(31)(26)sin27

Area 183.0units 2

Return to problems.